A body of mass m kg starts falling from a point 2R above the earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the earth’s surface [R-Radius of earth, M-Mass of earth, G-Gravitational constant]

Question

A body of mass m kg starts falling from a point 2R above the earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the earth’s surface [R-Radius of earth, M-Mass of earth, G-Gravitational constant] (A) (1/2) (GMm/R) (B) (1/6) (GMm/R) (C) (2/3) (GMm/R) (D) (1/3) (GMm/R)

Tardigrade Admin · Accepted Answer

When body starts falling toward earthâs surface its potential energy decreases so kinetic energy increases. Increase in kinetic energy = Decrease in potential energy Final kinetic energy - Initial kinetic energy = Initial potential energy - Final potential energy Final kinetic energy - 0 = (-(GMm/r1)) - (-(GMm/r2)) ∴ Final kinetic energy = ((-GMm/R +h1)) - (-(GMm/R +h2)) =(-(GMm/R+2R))-(-(GMm/R+R)) = - (GMm/3R) + (GMm/2R) = (1/6) (GMm/R)

Q. A body of mass $m k g$ starts falling from a point $2 R$ above the earth’s surface. Its kinetic energy when it has fallen to a point ‘ $R$ ’ above the earth’s surface [ $R$ -Radius of earth, $M$ -Mass of earth, $G$ -Gravitational constant]

$\frac{1}{2} \frac{GM m}{R}$

$\frac{1}{6} \frac{GM m}{R}$

$\frac{2}{3} \frac{GM m}{R}$

$\frac{1}{3} \frac{GM m}{R}$

Q. A body of mass mkg starts falling from a point 2R above the earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the earth’s surface [R-Radius of earth, M-Mass of earth, G-Gravitational constant]

21​RGMm​

61​RGMm​

32​RGMm​

31​RGMm​

Q. A body of mass $m k g$ starts falling from a point $2 R$ above the earth’s surface. Its kinetic energy when it has fallen to a point ‘ $R$ ’ above the earth’s surface [ $R$ -Radius of earth, $M$ -Mass of earth, $G$ -Gravitational constant]

$\frac{1}{2} \frac{GM m}{R}$

$\frac{1}{6} \frac{GM m}{R}$

$\frac{2}{3} \frac{GM m}{R}$

$\frac{1}{3} \frac{GM m}{R}$